Problem: Simplify; express your answer in exponential form. Assume $z\neq 0, r\neq 0$. $\dfrac{{(z^{2}r^{-4})^{3}}}{{(z^{4}r^{-1})^{-3}}}$
Explanation: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(z^{2}r^{-4})^{3} = (z^{2})^{3}(r^{-4})^{3}}$ On the left, we have ${z^{2}}$ to the exponent ${3}$ . Now ${2 \times 3 = 6}$ , so ${(z^{2})^{3} = z^{6}}$ Apply the ideas above to simplify the equation. $\dfrac{{(z^{2}r^{-4})^{3}}}{{(z^{4}r^{-1})^{-3}}} = \dfrac{{z^{6}r^{-12}}}{{z^{-12}r^{3}}}$ Break up the equation by variable and simplify. $\dfrac{{z^{6}r^{-12}}}{{z^{-12}r^{3}}} = \dfrac{{z^{6}}}{{z^{-12}}} \cdot \dfrac{{r^{-12}}}{{r^{3}}} = z^{{6} - {(-12)}} \cdot r^{{-12} - {3}} = z^{18}r^{-15}$